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An interesting Geometry problem

Let \(ABC\) be a triangle such that \(AB=AC\). Let \(M\) be a point on \(AB\) such that \(MA=MC\) and let \(N\) be a point on \(AC\) such that \(CN=CB\) and \(\angle BAC:\angle NBA=2:3\). Calculate the size of \(\angle NMC\).

You've found \(\angle BAC = 20^{\circ}\), but the problem is asking for \(\angle NMC\). Actually, after your solution this problem becomes literally equivalent to the one presented in Moscow 1952 and even has a name Langley's Adventitious Angles (an animated explanation here).